A lake is stocked with 545 fish of a new variety. The size of the lake, the availability of food, and the number of other fish restrict growth in the lake to a limiting value of 3406. The population of fish in the lake after time t, in
Months, is given by the function P(t) $P ( t ) = \frac { 3406 } { 1 + 5.07 e ^ { - 0.25 t } }$ . After how many months will the population be 2405?
A) 7 mo
B) 12 mo
C) 17 mo
D) 10 mo

Question

Quizplus · Accepted Answer

To find the time $t$ when the population $P(t)$ is 2405, we substitute 2405 for $P(t)$ in the given equation and solve for $t$:$$2405 = \frac{3406}{1 + 5.07e^{-0.25t}}$$Multiplying both sides by $1 + 5.07e^{-0.25t}$ and dividing both sides by 2405 gives:$$1 + 5.07e^{-0.25t} = \frac{3406}{2405}$$Solving for $e^{-0.25t}$ and then for $t$ will give the time in months. Following through the algebraic manipulations and solving for $t$ yields $t = 10$ months.

A Lake Is Stocked with 545 Fish of a New P(t)=34061+5.07e−0.25tP ( t ) = \frac { 3406 } { 1 + 5.07 e ^ { - 0.25 t } }P(t)=1+5.07e−0.25t3406​

A Lake Is Stocked with 545 Fish of a New $P ( t ) = \frac { 3406 } { 1 + 5.07 e ^ { - 0.25 t } }$